叉乘+二分解決POJ 2398
Description
Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box to put his toys in. Unfortunately, Reza is rebellious and obeys his parents by simply throwing his toys into the box. All the toys get mixed up, and it is impossible for Reza to find his favorite toys anymore.
Reza's parents came up with the following idea. They put cardboard partitions into the box. Even if Reza keeps throwing his toys into the box, at least toys that get thrown into different partitions stay separate. The box looks like this from the top:
We want for each positive integer t, such that there exists a partition with t toys, determine how many partitions have t, toys.
Reza's parents came up with the following idea. They put cardboard partitions into the box. Even if Reza keeps throwing his toys into the box, at least toys that get thrown into different partitions stay separate. The box looks like this from the top:
We want for each positive integer t, such that there exists a partition with t toys, determine how many partitions have t, toys.
Input
The input consists of a number of cases. The first line consists of six integers n, m, x1, y1, x2, y2. The number of cardboards to form the partitions is n (0 < n <= 1000) and the number of toys is given in m (0 < m <= 1000). The coordinates of the upper-left corner and the lower-right corner of the box are (x1, y1) and (x2, y2), respectively. The following n lines each consists of two integers Ui Li, indicating that the ends of the ith cardboard is at the coordinates (Ui, y1) and (Li, y2). You may assume that the cardboards do not intersect with each other. The next m lines each consists of two integers Xi Yi specifying where the ith toy has landed in the box. You may assume that no toy will land on a cardboard.
A line consisting of a single 0 terminates the input.
A line consisting of a single 0 terminates the input.
Output
For each box, first provide a header stating "Box" on a line of its own. After that, there will be one line of output per count (t > 0) of toys in a partition. The value t will be followed by a colon and a space, followed the number of partitions containing t toys. Output will be sorted in ascending order of t for each box.
Sample Input
4 10 0 10 100 0
20 20
80 80
60 60
40 40
5 10
15 10
95 10
25 10
65 10
75 10
35 10
45 10
55 10
85 10
5 6 0 10 60 0
4 3
15 30
3 1
6 8
10 10
2 1
2 8
1 5
5 5
40 10
7 9
0
Sample Output
Box
2: 5
Box
1: 4
2: 1
不寫不知道，一寫嚇一跳，原來這個是兩碼子的事。
中間有許多注意的東西，只有親自寫了，才知道的。
廢話不多說，直接上代碼。
View Code
#include "iostream" #include "algorithm" using namespace std; struct LineP { double up_x, down_x; }LP[1005]; bool Check(LineP p1, LineP p2) //排序子函數 { if(p1.up_x < p2.up_x) return true; return false; } double Judge(int j, double x, double y, double y1, double y2) //這個就是叉乘了 { return (LP[j].down_x - LP[j].up_x)*(y-y2) - (y2-y1)*(x-LP[j].down_x); } int main() { int n, m; double x1, y1, x2, y2, x, y; int ans[1005], realans[1005], low, high, mid; while(cin>>n && n) { memset(ans, 0, sizeof(ans)); memset(realans, 0, sizeof(realans)); cin>>m>>x1>>y1>>x2>>y2; for(int i=1; i<=n; i++) cin>>LP[i].up_x>>LP[i].down_x; LP[0].up_x = x1; LP[0].down_x = x1; LP[n+1].up_x = x2; LP[n+1].down_x = x2; sort(LP, LP+n+1, Check); for(int i=0; i<m; i++) { cin>>x>>y; low = 0; high = n+1; mid = (low+high)/2; if(Judge(0, x, y, y1, y2)==0) mid = 0; else if(Judge(n+1, x, y, y1, y2)==0) mid = n; else { while(low<high) { if(Judge(mid, x, y, y1, y2)<0) high = mid-1; else low = mid+1; mid = (low+high)/2; } if(Judge(mid, x, y, y1, y2) < 0) mid--; } ans[mid]++; } cout<<"Box"<<endl; for(int i=1; i<=n; i++) for(int j=0; j<=n+1; j++) if(ans[j]==i) realans[i]++; for(int i=0; i<=n; i++) if(realans[i]) cout<<i<<": "<<realans[i]<<endl; } }
posted on 2011-10-21 19:41 More study needed. 閱讀(298) 評論(0) 收藏 舉報
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